Arithmetic Sequences and Sums

Tutorial on sequences and summations.

A - Arithmetic Sequences

An arithmetic sequence is a sequence of numbers that is obtained by adding a constant number to the preceding number. The constant number is called the common difference.

Example 1:

0,6, 12, 18, 24, ... each term of the sequence is obtained by adding 6 to the preceding term.

If a1 is the first term and d is the common difference, the nth term an is given by

an = a1 + (n - 1)d

The sum Sn of the first n terms of an arithmetic sequence is given by

Sn = (n/2)(a1 + an) = (n/2)[2 a1 + (n - 1)d]



Examples with Detailed Solutions

Example 2:

Which term of the arithmetic sequence 2, 5, 8... is equal to 227?

Solution to Example 2:

The first term a
1 is 2 and the common difference is equal to: 5 - 2 = 8 - 5 = 3

Hence using the formula for the nth term, a
n = a1 + (n - 1)d to the term equal to 227, we can write the equation:

227 = 2 + (n - 1)3

Solve the above for n

n - 1 = (227 - 2) / 3 = 75 and n = 76

The 76th term is equal to 227.


Example 3:

How many consecutive odd integers of an arithmetic sequence, starting from 9, must be added in order to obtain a sum of 15,860?

Solution to Example 3:

The first term a
1 = 9 and d = 2 (the difference between any two consecutive odd integers). Hence the sum Sn of the n terms may be written as follows

S
n = (n/2)[2*a1 + (n - 1)d] = 15,960

With a
1 = 9 and d=2, the above equation in n may be written as follows

n
2 + 8 n - 15860 = 0

Solve the above for n

n = 122 and n = -130

The solution to the problem is that 122 consecutive odd numbers must be added in order to obtain a sum of 15,860.


Example 4:

What are the first 3 terms of an arithmetic sequence whose sum of the first n terms is equal to 2 n
2 + 5 n?

Solution to Example 4:

The sum S
n of the first n terms is given by Sn = 2 n2 + 5 n. The sum of the first n-1 terms is given by Sn-1 = 2(n-1)2 + 5(n-1). The difference between Sn and Sn-1 gives the nth term. Hence

a
n = Sn - Sn-1 = 2 n2 + 5 n - 2(n-1)2 - 5(n-1) = 2 n2 + 5 n - 2 n2 + 4 n - 2 - 5 n + 5

Which after simplification gives a
n = = 4 n + 3

We now use the formula for a
n to find the first 3 terms a1, a2 and a3.

a
1 = 4(1)+3 = 7 , a2= 4(2)+3 = 11 and a3 = 4(3)+3 = 15.

Example 5:

The sum of three consecutive numbers in an arithmetic sequence is equal 27 and their product is equal to 585. What are the three numbers?

Solution to Example 5:

Let the three numbers be x, x+d and x+3d where d is the common difference. Their sum is 27. Hence

x + (x+d) + (x+2d) = 27

Which gives 3x + 3d = 27 or x + d = 9

Their product is equal to 585. Hence

x(x+d)(x+2d) = 585

The equation x + d = 9 may be written as x = 9 - d

Substitute x by 9 - d in the equation x(x+d)(x+2d) = 585 to obtain

(9-d)(9)(9+d) = 585

Simplify to obtain

9
2 - d2 = 65

which gives d
2 = 16

Solve to obtain d = 4 or d = -4

Use x = 9 - d to obtain x = 5 for d = 4 and x = 13 for d = -4

There are two solutions to the given problem

1) d = 4 and the three terms are: 5, 9 and 13

2) d = -4 and the three terms are: 13, 9 and 5


Example 6:

The first three terms of an arithmetic sequence are as follows: x , 5 x/4 , 9/2. Find the first term x and the common difference d.

Solution to Example 6:

The common difference d is given by:

d = 5 x/4 - x = 9/2 - 5 x/4

Which gives 5 x/4 - x = 9/2 - 5 x/4

Solve the above equation for x

Common denominator: 5x/4 - 4x/4 = 18/2 - 5x/4

The equation: 6x = 18 and x = 3

d = 5 x / 4 - x = 15/4 - 3 = 3/4


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Updated: 21 October 2014 (A Dendane)

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